Worst Value-at-Risk under Known Margins

Marius Hofert

2018-04-28

library(qrmtools)
library(copula)
library(combinat)
library(sfsmisc)
doPDF <- FALSE

1 Homogeneous case

We start by considering the following setup in the homogeneous case, i.e., when all marginal distributions are equal.

qF <- function(p, th = 2) qPar(p, shape = th) # Pareto quantile function
pF <- function(q, th = 2) pPar(q, shape = th) # Pareto distribution function
dim <- 8 # variable dimension (we use 8 or 100 here)

1.1 Checks for method = “dual”

Investigate helper function \(h(s,t)\) (the function for the inner root-finding to compute \(D(s)\); see dual_bound()).

d <- 8 # dimension
s <- c(1, 5, 10, 100, 500, 1000)
t <- sapply(seq_along(s), function(i) {
    res <- exp(seq(log(1e-3), log(s[i]/d), length.out = 257))
    res[length(res)] <- s[i]/d # to avoid numerical issues (t > s/d)
    res
})
f <- sapply(seq_along(s), function(i)
            sapply(t[,i], function(t.)
                   qrmtools:::dual_bound_2(s[i], t = t., d = d, pF = pF) -
                   qrmtools:::dual_bound_2_deriv_term(s[i], t = t., d = d, pF = pF)))
palette <- colorRampPalette(c("maroon3", "darkorange2", "royalblue3"), space = "Lab")
cols <- palette(6)
if(doPDF)
    pdf(file = (file <- paste0("fig_worst_VaR_hom_dual_h_Par=2_d=",d,".pdf")),
        width = 6, height = 6)
plot(t[,1], f[,1], type = "l", log = "x", xlim = range(t), ylim = range(f), col = cols[1],
     xlab = "t", ylab = expression("h(s,t) for d = 8 and F being Par(2)"))
lines(t[,2], f[,2], col = cols[2])
lines(t[,3], f[,3], col = cols[3])
lines(t[,4], f[,4], col = cols[4])
lines(t[,5], f[,5], col = cols[5])
lines(t[,6], f[,6], col = cols[6])
abline(h = 0, lty = 2)
legend("topright", lty = rep(1,6), col = cols,
       bty = "n", legend = as.expression(lapply(1:6,
           function(i) substitute(s==s., list(s. = s[i])))))
if(doPDF) dev.off()

As we know, \(h(s,s/d) = 0\). We also see that \(s\) has to be sufficiently large in order to find a root \(h(s,t) = 0\) for \(t < s/d\).

Now let’s plot the dual bound \(D(s)\) for various \(\theta\) (this checks the outer root-finding).

theta <- c(0.5, 1, 2, 4) # theta values
s <- seq(48, 2000, length.out = 257) # s values
D <- sapply(theta, function(th)
            sapply(s, function(s.)
                   dual_bound(s., d = d, pF = function(q) pPar(q, shape = th)))) # (s, theta) matrix
if(doPDF)
    pdf(file = (file <- paste0("fig_worst_VaR_hom_dual_D_s_Par=",
                             paste0(theta, collapse="_"),"_d=",d,".pdf")),
        width = 6, height = 6)
plot(s, D[,1], type = "l", ylim = range(D), col = "maroon3",
     ylab = substitute("Dual bound D(s) for d ="~d.~"Par("*theta*") margins", list(d. = d)))
lines(s, D[,2], col = "darkorange2")
lines(s, D[,3], col = "royalblue3")
lines(s, D[,4], col = "black")
legend("topright", lty = rep(1,4),
       col = c("maroon3", "darkorange2", "royalblue3", "black"),
       bty = "n", legend = as.expression(lapply(1:4,
           function(j) substitute(theta==j, list(j = theta[j])))))
if(doPDF) dev.off()

1.2 Checks for method = “Wang”/“Wang.Par”

1.2.1 Check of auxiliary functions with numerical integration (for \(\theta = 2\))

Check Wang_h_aux().

d <- 8 # dimension
alpha <- 0.99 # confidence level
c <- seq(0, (1-alpha)/d, length.out = 129) # domain of h
h.aux <- qrmtools:::Wang_h_aux(c, level = alpha, d = d, qF = qF)
par(mar = c(5, 4+1, 4, 2) + 0.1) # increase space (for y axis label)
plot(c, h.aux, type = "l", xlab = "c (in initial interval)",
     ylab = expression(frac(d-1,d)~{F^{-1}}(a[c])+frac(1,d)~{F^{-1}}(b[c])))

Check the objective function \(h(c)\) (i.e., Wang_h() with numerical integration) on its domain. Note that \(h\) is used to determine the root in the open interval \((0,(1-\alpha)/d)\) for computing worst Value-at-Risk.

h <- sapply(c, function(c.) qrmtools:::Wang_h(c., level = alpha, d = d, qF = qF))
if(doPDF)
    pdf(file = (file <- paste0("fig_worst_VaR_",alpha,"_hom_Wang_h_Par=2_d=",d,"_num.pdf")),
        width = 6, height = 6)
plot(c, h, type = "l", xlab = "c (in initial interval)",
     ylab = substitute("h(c) for"~~alpha~"= 0.99 and d ="~d.~"Par(2) margins",
                       list(d. = d)))
abline(h = 0, lty = 2)
if(doPDF) dev.off()

Check the objective function \(h\) at the endpoints of its domain.

sapply(c(0, (1-alpha)/d), function(c.)
       qrmtools:::Wang_h(c., level = alpha, d = d, qF = qF)) # -Inf, 0
## [1] -Inf    0

\(-\infty\) is not a problem for root finding (for \(\theta > 1\); for \(\theta <= 1\) it is NaN, see below, and thus a problem!), but the \(0\) at the right endpoint is a problem.

method <- "Wang.Par" # this also holds for (the numerical) method = "Wang"
th <- 0.99
qrmtools:::Wang_h(0, level = alpha, d = d, method = method, shape = th) # NaN => uniroot() fails
## [1] NaN
## Note: Wang_h() is actually already NaN for c <= 1e-17
qrmtools:::Wang_h_aux(0, level = alpha, d = d, method = method, shape = th) # Inf
## [1] Inf

A proper initial interval \([c_l,u_l]\) with \(0<c_l<=c_u<(1-\alpha)/d\) (containing the root) is thus required. We have derived this in Hofert et al. (2017, ``Improved Algorithms for Computing Worst Value-at-Risk’’)).

1.2.2 Check of \(h(c)\) without numerical integration (for a range of \(\theta\))

Check the objective function \(h(c)\) (Wang_h() without numerical integration)

d <- dim # dimension
alpha <- 0.99 # confidence level
theta <- c(0.1, 0.5, 1, 5, 10, 50) # theta values
palette <- colorRampPalette(c("darkorange2", "maroon3", "royalblue3", "black"), space = "Lab")
cols <- palette(length(theta))
c <- seq(0, (1-alpha)/d, length.out = 2^13+1)
## => They all go further down to 0 if length.out is increased.
##    Smaller theta thus corresponds to a larger derivative in the root
##    Root-finding thus requires higher precision for smaller theta
h <- matrix(, nrow = length(c), ncol = length(theta))
for(j in 1:length(theta))
    h[,j] <- sapply(c, function(c.)
        qrmtools:::Wang_h(c., level = alpha, d = d, method = "Wang.Par", shape = theta[j]))
z <- h
z[z <= 0] <- NA # > 0 => makes log-scale possible
if(doPDF)
    pdf(file = (file <- paste0("fig_worst_VaR_",alpha,"_hom_Wang_h_Par_d=",d,".pdf")),
        width = 6, height = 6)
plot(NA, xlim = range(c), ylim = range(z, na.rm = TRUE), log = "y", yaxt = "n", xlab = "c",
     ylab = substitute("h(c) for"~~alpha~"= 0.99, d ="~d.~"Par("*theta*") margins",
                       list(d. = d)))
eaxis(2)
for(k in 1:length(theta))
    lines(c, z[,k], col = cols[k])
legend("topleft", bty = "n", lty = rep(1,length(theta)), col = cols,
       legend = as.expression(lapply(1:length(theta),
       function(k) substitute(theta==k, list(k = theta[k])))))
if(doPDF) dev.off()

1.3 Compute best/worst \(\mathrm{VaR}_\alpha\) (via “Wang.Par”)

After dealing with various numerical issues, we can now look at some example calculations of best/worst Value-at-Risk. We first plot Value-at-Risk as a function in \(\alpha\).

d <- dim # dimension
alpha <- 1-2^seq(-0.001, -10, length.out = 128) # confidence levels; concentrated near 1
theta <- c(0.1, 0.5, 1, 5, 10, 50) # theta values
VaR <- simplify2array(sapply(alpha, function(a)
    sapply(theta, function(th) VaR_bounds_hom(a, d = d, method = "Wang.Par",
                                              shape = th)), simplify = FALSE))
## => (best/worst VaR, theta, alpha)-matrix
if(doPDF)
    pdf(file = (file <- paste0("fig_worst_VaR_hom_Wang_Par_d=",d,".pdf")),
        width = 7, height = 7)
par(mar = c(5, 4+1, 4, 2) + 0.1) # increase space (for y axis label)
plot(NA, xlim = range(alpha), ylim = range(VaR), log = "xy", yaxt = "n",
     xlab = expression(1-alpha),
     ylab = as.expression(substitute(underline(VaR)[alpha](L^{"+"})~"(dashed) and"~bar(VaR)[alpha](L^{"+"})~
                                     "(solid) for d ="~d.~"and Par("*theta*") margins",
                                     list(d. = d))))
eaxis(2)
for(k in 1:length(theta)) {
    lines(1-alpha, VaR[2,k,], col = cols[k]) # worst VaR
    lines(1-alpha, VaR[1,k,], col = cols[k], lty = 2) # best VaR
}
legend("topright", bty = "n", lty = rep(1,length(theta)), col = cols,
       legend = as.expression(lapply(1:length(theta),
       function(k) substitute(theta==k, list(k = theta[k])))))
if(doPDF) dev.off()

Now consider best/worst Value-at-Risk as a function in \(d\).

d <- seq(2, 1002, by = 20) # dimensions
alpha <- 0.99 # confidence level
theta <- c(0.1, 0.5, 1, 5, 10, 50) # theta values
VaR <- simplify2array(sapply(d, function(d.)
    sapply(theta, function(th) VaR_bounds_hom(alpha, d = d., method = "Wang.Par",
                                              shape = th)), simplify = FALSE))
## => (best/worst VaR, theta, d)-matrix
if(doPDF)
    pdf(file = (file <- paste0("fig_worst_VaR_hom_Wang_Par_alpha=",alpha,"_in_d.pdf")),
        width = 7, height = 7)
par(mar = c(5, 4+1, 4, 2) + 0.1) # increase space (for y axis label)
plot(NA, xlim = range(d), ylim = range(VaR), log = "xy", yaxt = "n",
     xlab = expression(d),
     ylab = as.expression(substitute(underline(VaR)[a](L^{"+"})~"(dashed) and"~bar(VaR)[a](L^{"+"})~
                                     "(solid) for Par("*theta*") margins",
                                     list(a = alpha))))
eaxis(2)
for(k in 1:length(theta)) {
    lines(d, VaR[2,k,], col = cols[k]) # worst VaR
    lines(d, VaR[1,k,], col = cols[k], lty = 2) # best VaR
}
legend("topleft", bty = "n", lty = rep(1,length(theta)), col = cols,
       legend = as.expression(lapply(1:length(theta),
       function(k) substitute(theta==k, list(k = theta[k])))))
if(doPDF) dev.off()

We can also consider best/worst Value-at-Risk as a function in \(\theta\). Note that, as before, best Value-at-Risk is the same for all \(d\) here (depicted in a green dashed line below).

d <- c(2, 10, 20, 100, 200, 1000) # dimensions
theta <- 10^seq(-1, log(50, base = 10), length.out = 50) # theta values
alpha <- 0.99 # confidence level
VaR <- simplify2array(sapply(d, function(d.)
    sapply(theta, function(th) VaR_bounds_hom(alpha, d = d., method = "Wang.Par",
                                              shape = th)), simplify = FALSE))
## => (best/worst VaR, theta, d)-matrix
if(doPDF)
    pdf(file = (file <- paste0("fig_worst_VaR_hom_Wang_Par_alpha=",alpha,"_in_theta.pdf")),
        width = 6, height = 6)
par(mar = c(5, 4+1, 4, 2) + 0.1) # increase space (for y axis label)
plot(NA, xlim = range(theta), ylim = range(VaR), log = "xy", yaxt = "n",
     xlab = expression(theta),
     ylab = as.expression(substitute(underline(VaR)[a](L^{"+"})~"(dashed) and"~
                          bar(VaR)[a](L^{"+"})~"(solid) for Par("*theta*") margins",
                          list(a = alpha))))
eaxis(2)
bestVaR <- VaR[1,,1]
for(k in 1:length(d)) {
    lines(theta, VaR[2,,k], col = cols[k]) # worst VaR
    if(k >= 2) stopifnot(all.equal(bestVaR, VaR[1,,k]))
}
lines(theta, bestVaR, col = "darkgreen", lty = 2) # best VaR
legend("topright", bty = "n", lty = rep(1,length(d)), col = cols,
       legend = as.expression(lapply(1:length(d),
       function(k) substitute(d==k, list(k = d[k])))))
if(doPDF) dev.off()

1.4 Comparison between various methods for computing worst Value-at-Risk

Now consider a graphical comparison between the various methods. To stress the numerical challenges, let us include the following implementation of Wang’s approach (with various switches to highlight the individual numerical challenges) for the Pareto case.

## Initial interval for the root finding in case of worst VaR
init_interval <- function(alpha, d, shape, trafo = FALSE, adjusted = FALSE)
{
    if(trafo) {
        low <- if(shape == 1) {
            d/2
        } else {
            (d-1)*(1+shape)/(d-1+shape)
        }
        up <- if(shape > 1) {
            r <- (1+d/(shape-1))^shape
            if(adjusted) 2*r else r
        } else if(shape == 1) {
            e <- exp(1)
            (d+1)^(e/(e-1))
        } else {
            d*shape/(1-shape)+1
        }
        c(low, up)
    } else {
        low <- if(shape > 1) {
            r <- (1-alpha)/((d/(shape-1)+1)^shape + d-1)
            if(adjusted) r/2 else r
        } else if(shape == 1) {
            e <- exp(1)
            (1-alpha)/((d+1)^(e/(e-1))+d-1)
        } else {
            r <- (1-shape)*(1-alpha)/d
            if(adjusted) r/2 else r
        }
        up <- if(shape == 1) (1-alpha)/(3*d/2-1)
              else (1-alpha)*(d-1+shape)/((d-1)*(2*shape+d))
        c(low, up)
    }
}

## Function to compute the best/worst Value-at-Risk in the homogeneous case with
## Par(theta) margins
VaR_hom_Par <- function(alpha, d, shape, method = c("worst", "best"),
                        trafo = FALSE, interval = NULL, adjusted = FALSE,
                        avoid.cancellation = FALSE, ...)
{
    ## Pareto quantile function
    qF <- function(p) (1 - p)^(-1/shape) - 1

    ## Compute \bar{I}
    Ibar <- function(a, b, alpha, d, shape)
    {
        if(shape == 1) log((1-a)/(1-b))/(b-a) - 1
        else (shape/(1-shape))*((1-b)^(1-1/shape)-(1-a)^(1-1/shape))/(b-a) - 1
    }

    ## Main
    method <- match.arg(method)
    switch(method,
    "worst" = {

        ## Distinguish according to whether we optimize the auxiliary function
        ## on a transformed scale
        h <- if(trafo) {
            ## Auxiliary function to find the root of on (1, Inf)
            if(shape == 1) {
                function(x) x^2 + x*(-d*log(x)+d-2)-(d-1)
            } else {
                function(x)
                (d/(1-shape)-1)*x^(-1/shape + 1) - (d-1)*x^(-1/shape) + x - (d*shape/(1-shape) + 1)
            }
        } else {
            ## Auxiliary function to find the root of on (0, (1-alpha)/d)
            function(c) {
                a <- alpha+(d-1)*c
                b <- 1-c
                Ib <- if(c == (1-alpha)/d) { # Properly deal with limit c = (1-alpha)/d
                    ((1-alpha)/d)^(-1/shape) - 1
                } else {
                    Ibar(a = a, b = b, alpha = alpha, d = d, shape = shape)
                }
                Ib - (qF(a)*(d-1)/d + qF(b)/d)
            }
        }

        ## Do the optimization
        if(is.null(interval)) interval <- init_interval(alpha, d, shape,
                                                        trafo = trafo, adjusted = adjusted)
        c <- uniroot(h, interval = interval, ...)$root
        if(trafo) # convert value back to the right scale (c-scale)
            c <- (1-alpha)/(c+d-1)
        if(avoid.cancellation) {
            t1 <- (1-alpha)/c-(d-1)
            d * ((c^(-1/shape)/d) * ((d-1)*t1^(-1/shape) + 1) - 1) # = qF(a)*(d-1) + qF(b)
        } else {
            a <- alpha+(d-1)*c
            b <- 1-c
            qF(a)*(d-1) + qF(b)
        }
    },
    "best" = {
        max((d-1)*0 + (1-alpha)^(-1/shape)-1, # Note: Typo in Wang, Peng, Yang (2013)
            d*Ibar(a = 0, b = alpha, alpha = alpha, d = d, shape))
    },
    stop("Wrong 'method'"))
}

For the comparison, consider the following setup.

alpha <- 0.99 # confidence level
d <- dim # dimension
n.th <- 32 # number of thetas
th <- seq(0.2, 5, length.out = n.th) # thetas
qFs <- lapply(th, function(th.) {th.; function(p) qPar(p, shape = th.)}) # n.th-vector of Pareto quantile functions
pFs <- lapply(th, function(th.) {th.; function(q) pPar(q, shape = th.)}) # n.th-vector of Pareto dfs
N <- 1e4 # number of discretization points for RA(); N = 1e5 does not improve the situation

Now compute the values and plot them.

res <- matrix(, nrow = n.th, ncol = 7)
colnames(res) <- c("Wang", "straightforward", "transformed", "Wang.Par",
                   "dual", "RA.low", "RA.up")
pb <- txtProgressBar(max = n.th, style = if(isatty(stdout())) 3 else 1) # setup progress bar
on.exit(close(pb)) # on exit, close progress bar
for(i in seq_len(n.th)) {
    ## "Wang" (numerical integration with smaller uniroot() tolerance; still
    ## numerically critical -- we catch "the integral is probably divergent"-errors here)
    Wang.num.res <- tryCatch(VaR_bounds_hom(alpha, d = d, qF = qFs[[i]])[2], error = function(e) e)
    res[i,"Wang"] <- if(is(Wang.num.res, "simpleError")) NA else Wang.num.res
    ## Our straightforward implementation
    res[i,"straightforward"] <- VaR_hom_Par(alpha, d = d, shape = th[i])
    ## Our straightforward implementation based on the transformed auxiliary function
    res[i,"transformed"] <- VaR_hom_Par(alpha, d = d, shape = th[i], trafo = TRUE)
    ## "Wang.Par" (using a smaller uniroot() tolerance and adjusted initial interval)
    res[i,"Wang.Par"] <- VaR_bounds_hom(alpha, d = d, method = "Wang.Par", shape = th[i])[2]
    ## "dual" (with uniroot()'s default tolerance)
    res[i,"dual"] <- VaR_bounds_hom(alpha, d = d, method = "dual",
                                    interval = crude_VaR_bounds(alpha, qF = qFs[[i]], d = d),
                                    pF = pFs[[i]])[2]
    ## Rearrangement Algorithm
    set.seed(271) # use the same random permutation for each theta
    RA. <- RA(alpha, qF = rep(qFs[i], d), N = N)
    res[i,"RA.low"] <- RA.$bounds[1]
    res[i,"RA.up"]  <- RA.$bounds[2]
    ## Progress
    setTxtProgressBar(pb, i) # update progress bar
}
res. <- res/res[,"dual"] # standardize (by dual method)
ylim <- range(res., na.rm = TRUE)
if(doPDF)
    pdf(file = (file <- paste0("fig_worst_VaR_",alpha,"_hom_comparison_d=",
                             d,"_N=",N,".pdf")), width = 7, height = 7)
par(mar = c(5, 4+1, 4, 2) + 0.1) # increase space (for y axis label)
plot(th, res.[,"Wang"], type = "l", ylim = ylim,
     xlab = expression(theta), ylab = substitute("Standardized (by dual method)"~
     bar(VaR)[0.99]~"for d ="~d.~"Par("*theta*") margins", list(d. = d)),
     col = "gray60", lty = 2, lwd = 5.5) # Wang (with numerical integration)
lines(th, res.[,"straightforward"], col = "maroon3", lty = 1, lwd = 1) # still bad (although we have an initial interval)
lines(th, res.[,"transformed"], col = "black", lty = 1, lwd = 1) # okay
lines(th, res.[,"Wang.Par"], col = "royalblue3", lty = 2, lwd = 2.5) # Wang Pareto (wo num. integration)
lines(th, res.[,"RA.low"], col = "black", lty = 3, lwd = 1) # lower RA bound
lines(th, res.[,"RA.up"],  col = "black", lty = 2, lwd = 1) # upper RA bound
legend("topright", bty = "n",
       col = c("gray60", "maroon3", "black", "royalblue3", "black", "black"),
       lty = c(2,1,1,2,3,2), lwd = c(5.5,1,1,2.5,1,1),
       legend = c("Wang (num. int.)", "Wang Pareto (straightforward)",
                  "Wang Pareto (transformed)", "Wang Pareto (wo num. int.)",
                  "Lower RA bound", "Upper RA bound"))
if(doPDF) dev.off()

So what goes wrong with our straightforward implementation? Let’s start with the obvious. As we can infer from the x-axis scale in the plots of \(h\) above, uniroot()’s default tolerance tol = .Machine$double.eps^0.25(\(\approx 0.0001220703\) here) is too large to determine the root accurately. In fact, if we choose a smaller tol, we have no problem.

tol <- 2.2204e-16
wVaR.tol <- sapply(th, function(th.)
    VaR_hom_Par(alpha = alpha, d = d, shape = th., tol = tol))
plot(th, wVaR.tol/res[,"dual"], type = "l", ylim = ylim,
     xlab = expression(theta), ylab = "Wang's approach (straightforward) but with smaller tol")

So is this the solution to all the problems? No. Consider the setup as before, where \(d\) is running (we catch the appearing errors here).

alpha <- 0.99 # confidence level
d <- seq(2, 1002, by = 20) # dimensions
theta <- c(0.1, 0.5, 1, 5, 10, 50) # theta values
VaR <- simplify2array(sapply(d, function(d.)
    sapply(theta, function(th) {
        res <- tryCatch(VaR_hom_Par(alpha, d = d., shape = th, tol = tol),
                        error = function(e) e)
        if(is(res, "simpleError")) { warning(conditionMessage(res)); NA }
        else res
    }), simplify = FALSE))

By using warnings() we see that we cannot even determine the root in all cases as the function values are numerically not of opposite sign at the endpoints of the theoretically correct initial interval containing the root. We can solve this (non-elegantly) by simply adjusting the lower bound.

VaR <- simplify2array(sapply(d, function(d.)
    sapply(theta, function(th) VaR_hom_Par(alpha, d = d., shape = th, tol = tol, adjusted = TRUE)),
    simplify = FALSE))

So we see that we need a smaller tolerance and to extend the initial interval in order to get reasonable results.

What about using the transformed auxiliary function directly? Is this a numerically more stable approach?

d <- 500
th <- 20
VaR_hom_Par(alpha, d = d, shape = th, trafo = TRUE) # Inf
## [1] Inf

As we can see, there is a problem as well. To understand it, consider the following minimized version of VaR_hom_Par() (for computing worst Value-at-Risk for \(\theta\neq 1\) only).

h <- function(x)
     (d/(1-th)-1)*x^(-1/th + 1) - (d-1)*x^(-1/th) + x - (d*th/(1-th) + 1)
interval <- init_interval(alpha, d, th, trafo = TRUE)
x <- uniroot(h, interval = interval)$root
(c <- (1-alpha)/(x+d-1)) # convert back to c-scale
## [1] 1.869497e-31
a <- alpha+(d-1)*c
b <- 1-c
qPar(a, shape = th)*(d-1) + qPar(b, shape = th) # Inf
## [1] Inf
stopifnot(b == 1) # => b is 1 => qPar(b, shape = th) = Inf

As we can see, \(c>0\) is so small that \(b = 1-c\) is numerically equal to 1 and thus the Pareto quantile evaluated as \(\infty\). This cancellation can be avoided by simplifying the terms (note that qPar(1-c, shape = th) = c^(-1/th)-1).

qPar(a, shape = th)*(d-1) + c^(-1/th)-1
## [1] 162.5923

We can apply this to the variable combinations as before. Note that due to the partly extreme values of \(d\) and \(\theta\), we also need to adjust the range here in order for a root to be found.

d <- seq(2, 1002, by = 20) # dimensions
theta <- c(0.1, 0.5, 1, 5, 10, 50) # theta values
VaR. <- simplify2array(sapply(d, function(d.)
    sapply(theta, function(th) VaR_hom_Par(alpha, d = d., shape = th,
                                           trafo = TRUE, adjusted = TRUE,
                                           avoid.cancellation = TRUE)),
    simplify = FALSE))

We can now compare the values (based on the non-transformed/transformed auxiliary function) in a graph.

ylim <- range(VaR, VaR.)
if(doPDF)
    pdf(file = (file <- paste0("fig_worst_VaR_",alpha,"_hom_comparison_num_problems.pdf")),
        width = 7, height = 7)
par(mar = c(5, 4+1, 4, 2) + 0.1) # increase space (for y axis label)
plot(d, VaR[1,], type = "l", ylim = ylim, log = "y", yaxt = "n", xlab = "d",
     ylab = as.expression(substitute(bar(VaR)[a](L^{"+"})~
            "directly (solid) or with transformed h (dashed) for Par("*theta*") margins",
            list(a = alpha))), col = cols[1])
eaxis(2)
for(k in 2:length(theta)) lines(d, VaR [k,], col = cols[k])
for(k in 1:length(theta)) lines(d, VaR.[k,], col = cols[k], lty = 2, lwd = 2)
legend("right", bty = "n",
       col = cols, lty = rep(1, length(theta)),
       legend = as.expression( lapply(1:length(theta), function(k)
           substitute(theta==th, list(th = theta[k]))) ))
if(doPDF) dev.off()

We see that the results differ slightly if both \(d\) and \(\theta\) are large. It remains numerically challenging to appropriately compute worst Value-at-Risk in this case. However, both of the above approaches work for dimensions even as large as \(10^5\).

2 Inhomogeneous case

2.1 Run-time comparison (straightforward vs efficient implementation)

A straightforward (but inefficient) implementation of a basic rearrange(, sample = FALSE, is.sorted = TRUE) is the following. Note that our implementation in qrmtools allows for much greater functionality and is faster.

basic_rearrange_worst_VaR <- function(X, tol = 0)
{
    N <- nrow(X)
    d <- ncol(X)
    m.rs.old <- min(rowSums(X))
    while (TRUE) {
        Y <- X
        for(j in 1:d)
            Y[,j] <- sort(Y[,j], decreasing = TRUE)[rank(rowSums(Y[,-j, drop = FALSE]), ties.method = "first")]
        Y.rs <- rowSums(Y)
        m.rs.new <- min(Y.rs)
        tol. <- abs((m.rs.new - m.rs.old)/m.rs.old)
        tol.reached <- if(is.null(tol)) {
            identical(Y, X)
        } else { tol. <= tol }
        if(tol.reached) {
            break
        } else {
            m.rs.old <- m.rs.new
            X <- Y
        }
    }
    min(rowSums(Y))
}

We now compare this to the actual implementation (rearrange(, sample = FALSE, is.sorted = TRUE)). To this end, we consider the following setup.

## Build the input matrix (for worst VaR)
alpha <- 0.99 # confidence level
N <- 2^9 # 512
p <- alpha + (1-alpha)*0:(N-1)/N
if(FALSE)
    d <- 2^(4:10) # 16, ..., 1024
d <- 2^(4:8) # 16, ..., 256 (to save time here)
res <- matrix(, nrow = length(d), ncol = 2) # matrix containing run time results
colnames(res) <- c("basic", "sophisticated")
qF <- function(p, th = 2) qPar(p, shape = th) # Pareto quantile function

## For each d, measure the run time
for(i in seq_along(d)) {
    ## cat("Working on d =",d[i],"\n")
    X <- sapply(rep(list(qF), d[i]), function(qF.) qF.(p))
    res[i,] <- c(system.time(basic_rearrange_worst_VaR(X))[["elapsed"]],
                 system.time(rearrange(X, sample = FALSE, is.sorted = TRUE))[["elapsed"]])
}

Here’s a plot which shows the improvement in run time (in %). For larger dimensions \(d\), the improvement is very close to 100%.

if(doPDF)
    pdf(file = (file <- paste0("fig_ARA_speed-up.pdf")),
        width = 6, height = 6)
plot(d, (1-res[,2]/res[,1])*100, type = "b", log = "x", ylim = c(0,100),
     xlab = "d", ylab = "Relative speed-up (in %) of implemented ARA()")
if(doPDF) dev.off()

2.2 How rearrange() acts on specific matrices

To see how rearrange() actually proceeds, consider the following example. Due to trace = TRUE, the matrix is printed after each column rearrangement. A “|” and “=” indicate whether the respective column has changed or not, respectively, and the last two printed columns provide the row sums over all other columns but the current one as well as the new updated row sums (over all columns) after the rearrangement. We see that for tol = NULL the algorithm stops after the first time \(d\) (here: 3) consecutive rearrangements left the matrix unchanged.

A <- matrix(c(1:4, 2*(1:4)-1, 2^(0:3)), ncol = 3)
rearrange(A, tol = NULL, sample = FALSE, is.sorted = TRUE, trace = TRUE)
##           
## [1,] 1 1 1
## [2,] 2 3 2
## [3,] 3 5 4
## [4,] 4 7 8
##      |     -col sum
## [1,] 4 1 1    2   6
## [2,] 3 3 2    5   8
## [3,] 2 5 4    9  11
## [4,] 1 7 8   15  16
##        |   -col sum
## [1,] 4 5 1    5  10
## [2,] 3 7 2    5  12
## [3,] 2 3 4    6   9
## [4,] 1 1 8    9  10
##          | -col sum
## [1,] 4 5 2    9  11
## [2,] 3 7 1   10  11
## [3,] 2 3 4    5   9
## [4,] 1 1 8    2  10
##      |     -col sum
## [1,] 3 5 2    7  10
## [2,] 2 7 1    8  10
## [3,] 4 3 4    7  11
## [4,] 1 1 8    9  10
##        =   -col sum
## [1,] 3 5 2    5  10
## [2,] 2 7 1    3  10
## [3,] 4 3 4    8  11
## [4,] 1 1 8    9  10
##          = -col sum
## [1,] 3 5 2    8  10
## [2,] 2 7 1    9  10
## [3,] 4 3 4    7  11
## [4,] 1 1 8    2  10
##      =     -col sum
## [1,] 3 5 2    7  10
## [2,] 2 7 1    8  10
## [3,] 4 3 4    7  11
## [4,] 1 1 8    9  10
## $bound
## [1] 10
## 
## $tol
## [1] 0
## 
## $converged
## [1] TRUE
## 
## $opt.row.sums
## [1]  6  9  9 10 10 10 10
## 
## $X.rearranged
##      [,1] [,2] [,3]
## [1,]    3    5    2
## [2,]    2    7    1
## [3,]    4    3    4
## [4,]    1    1    8

This is the highest possible minimal row sum and thus the optimally rearranged matrix. Let’s consider another example.

B <- matrix(rep(1:3, 3), ncol = 3)
rearrange(B, tol = NULL, sample = FALSE, is.sorted = TRUE, trace = TRUE)
##           
## [1,] 1 1 1
## [2,] 2 2 2
## [3,] 3 3 3
##      |     -col sum
## [1,] 3 1 1    2   5
## [2,] 2 2 2    4   6
## [3,] 1 3 3    6   7
##        =   -col sum
## [1,] 3 1 1    4   5
## [2,] 2 2 2    4   6
## [3,] 1 3 3    4   7
##          = -col sum
## [1,] 3 1 1    4   5
## [2,] 2 2 2    4   6
## [3,] 1 3 3    4   7
##      =     -col sum
## [1,] 3 1 1    2   5
## [2,] 2 2 2    4   6
## [3,] 1 3 3    6   7
## $bound
## [1] 5
## 
## $tol
## [1] 0
## 
## $converged
## [1] TRUE
## 
## $opt.row.sums
## [1] 5 5 5 5
## 
## $X.rearranged
##      [,1] [,2] [,3]
## [1,]    3    1    1
## [2,]    2    2    2
## [3,]    1    3    3

Here we do not reach an optimal rearrangement (6 would be a better minimal row sum, attainable via the rearrangement matrix(c(1,2,3, 2,3,1, 3,1,2), ncol = 3)). It remains an open problem how an efficient algorithm can avoid such a case.

2.3 Convergence

Whether rearrange() converges or not depends to a large degree on the underlying sorting algorithm. We found that unstable sorting algorithms such as C’s qsort() lead to rearrange() not terminating on certain input matrices. One example is the above matrix B, for which the 2nd/3rd column are constantly swapped after the first column being rearranged since the row sum of all others is (4,4,4) and termination then totally depends on how ranks are assigned to these three equal numbers. The current implementation is also not based on a stable sorting algorithm but a) is comparably fast and b) terminates on all input matrices. To see the latter, consider the following example in which we investigate the possible output matrices of rearrange() on all input matrices (obviously only for \(d = 3\) due to run time; larger \(N\) can be chosen but note that the effort is \((N!)^{d-1}\)).

### Investigate the probability of certain rearranged matrices to appear #######

##' @title Create a list of all possible input matrices for d = 3
##' @param N The number of discretization points
##' @param method The method how to create the list
##' @return A N!^2-list (containing all possible (N,d) input matrices with 1:N in
##'         the first column)
##' @author Marius Hofert
create_mat <- function(N)
{
    x_perm <- permn(N) # N!-long list containing all possible permutations of 1:N
    N. <- factorial(N)
    lst <- vector("list", length = N.^2)
    cnt <- 0
    for (i in 1:N.) { # double 'for' not time critical here
        for (j in 1:N.){
            cnt <- cnt+1
            lst[[cnt]] <- cbind(1:N, x_perm[[i]], x_perm[[j]])
        }
    }
    lst
}


## Main
N <- 5 # chosen N (<= 6 due to extensive run time)
system.time(mat  <- create_mat(N)) # create the list of matrices (N!^2-many)
##    user  system elapsed 
##   0.026   0.000   0.026
## Rearrange all of them
N. <- factorial(N)^2
system.time(mat.ra <- lapply(1:N., function(i) {
                ## if(i %% (N./20) == 0) cat(round(100*i/N.),"% done!\n", sep = "");
                rearrange(mat[[i]], tol = NULL, sample = FALSE)$X.rearranged
            }))
##    user  system elapsed 
##   3.671   0.005   3.679
## Go through all of the rearranged matrices, 'uniquify' them and count
lst <- list(mat.ra[[1]])
freq <- c(1)
for(i in 2:length(mat.ra))
{
    bool <- sapply(1:length(lst),
                   function(i) { identical(lst[[i]], mat.ra[[i]]) }) # is mat.ra in lst?
    if(any(bool)) { # mat.ra has been observed before
        freq[min(which(bool))] <- freq[min(which(bool))] + 1
    } else { # mat.ra has not been observed before
        lst <- c(lst, mat.ra[[i]]) # append it to lst()
        freq <- c(freq, 1) # append its frequency
    }
}

## Result
lst # final rearranged matrices
## [[1]]
##      [,1] [,2] [,3]
## [1,]    5    1    1
## [2,]    4    2    2
## [3,]    3    3    3
## [4,]    2    4    4
## [5,]    1    5    5
freq/N. # their corresponding probability
## [1] 1

As we can see, not only does rearrange() terminate on all inputs, it also generates the same output matrix for all inputs.

2.4 A real data application

We now consider the task of computing best and worst Value-at-Risk (VaR) at confidence level 99.9% for a real data set. To illustrate the ideas, we simply consider the negative log-returns of the constituents of the SMI from 2011-09-12 to 2012-03-28. In a more realistic example where the marginal loss distributions typically come from different business lines or sub-units and there is less information about the dependence (or no hope to estimate it), one would consider real losses.

First, let’s compute the losses.

data(SMI.12)
L <- -apply(log(SMI.12), 2, diff) # more sophisticated methods are available
n <- nrow(L)
d <- ncol(L)

Now let’s fit GPDs to each margin, i.e., each SMI constituent. As threshold, we simply choose the 80% quantile for each of margins (more sophisticated methods are available).

res <- vector("list", length = d)
names(res) <- colnames(L)
for(k in seq_len(d)) {
    ## Determine the threshold for company k
    L. <- L[,k]
    u <- quantile(L., probs = 0.8, names = FALSE) # threshold

    ## Fit a GPD to the excesses
    excess <- L.[L.>u]-u # excesses
    fit <- fit_GPD_MLE(excess)
    stopifnot(fit$converged == 0)
    xi <- fit$par[["shape"]] # fitted xi
    beta <- fit$par[["scale"]] # fitted beta
    stopifnot(is.numeric(xi), is.numeric(beta))

    ## Graphical goodness-of-fit check for the GPD fit
    if(FALSE) {
        excess. <- sort(excess) # sorted data
        qF <- function(p) qGPD(p, shape = xi, scale = beta)
        qF. <- qF(ppoints(length(excess.))) # theoretical quantiles
        stock <- names(res)[k]
        plot(qF., excess., xlab = "Theoretical quantiles",
             ylab = "Sample quantiles", main = paste0("Q-Q plot for the fitted GPD(",
             round(xi, 2),", ",round(beta, 2),") distribution for ", stock))
        qqline(y = as.numeric(excess.), distribution = qF)
    }

    ## Update res
    res[[k]] <- list(loss = L., excess = excess, u = u, xi = xi, beta = beta)
}

We can now apply the (Adaptive) Rearrangement Algorithm.

xi. <- sapply(res, function(x) x$xi) # all fitted xi's
beta. <- sapply(res, function(x) x$beta) # all fitted beta's
alpha <- 0.99 # confidence level
qF <- lapply(res, function(r) { function(p) qGPD(p, shape = r$xi, scale = r$beta) }) # list quantile functions
set.seed(271) # set a seed (for reproducibility)
res.ARA <- ARA(alpha, qF = qF) # apply ARA()
stopifnot(res.ARA$converged) # check convergence
res.ARA$bounds # ARA bounds on worst VaR
##      low       up 
## 1.463087 1.472194
res.RA <- RA(alpha, qF = qF, N = res.ARA$N.used, abstol = 0) # apply RA() (with the same N)
stopifnot(res.RA$converged) # check convergence
res.RA$bounds # RA bounds on worst VaR
##      low       up 
## 1.463085 1.472188

To see how the bounds on worst VaR depend on the chosen tolerance, consider the following.

## Build the input matrices for rearrange()
N <- res.ARA$N.used
p.low <- alpha + (1-alpha)*(0:(N-1))/N # probability points for worst VaR (lower bound)
p.up  <- alpha + (1-alpha)*(1:N)/N # probability points for worst VaR (upper bound)
X.low <- sapply(qF, function(qF) qF(p.low)) # input matrix (lower bound)
X.up  <- sapply(qF, function(qF) qF(p.up)) # input matrix (upper bound)
X.up[N,] <- sapply(1:d, function(j) if(is.infinite(X.up[N,j]))
                   qF[[j]](alpha+(1-alpha)*(1-1/(2*N))) else X.up[N,j])

## Apply rearrange() for various tolerances (incl. NULL)
tol <- seq(0, 0.5, length.out = 21) # considered tolerances
res.low  <- sapply(tol, function(t) rearrange(X.low,  tol = t, sample = FALSE, is.sorted = TRUE)$bound)
res.low. <- rearrange(X.low, tol = NULL, sample = FALSE, is.sorted = TRUE)$bound # for tol = NULL
res.up  <- sapply(tol, function(t) rearrange(X.up, tol = t, sample = FALSE, is.sorted = TRUE)$bound)
res.up. <- rearrange(X.up, tol = NULL, sample = FALSE, is.sorted = TRUE)$bound # for tol = NULL
## Plot the lower and upper bound on worst VaR as a function in the chosen tol
if(doPDF)
    pdf(file = (file <- paste0("fig_worst_VaR_bounds_application.pdf")),
        width = 6, height = 6)
par(mar = c(5, 4+1, 4, 2) + 0.1) # increase space (for y axis label)
plot(tol, res.low, type = "b", log = "y", ylim = range(res.low, res.low., res.up, res.up.),
     col = "royalblue3", xlab = "Relative tolerance tol of rearrange()",
     ylab = substitute("Bounds on"~bar(VaR)[a](L^{"+"}), list(a = alpha)))
points(0, res.low., pch = 3, col = "royalblue3") # draw tol = NULL result at 0, too (as '+')
lines(tol, res.up, type = "b")
points(0, res.up., pch = 3) # draw tol = NULL result at 0, too (as '+')
legend("topright", bty = "n", lty = rep(1,2),
       col = c("black", "royalblue3"), legend = c("Upper bound", "Lower bound"))
if(doPDF) dev.off()

As we see, the results for tol = NULL (i.e., all columns oppositely ordered with respect to the sum of all others; indicated by ‘+’) are identical to the results for tol = 0 here; the latter are typically much faster to compute.